1/ 2 n2(g) + 1/2 o2(g) → no(g)
1/ 2 n2(g) + o2(g) → no2(g)
Chemistry, 26.11.2019 22:31 lilgunter12
Consider the reactions:
1/ 2 n2(g) + 1/2 o2(g) → no(g)
1/ 2 n2(g) + o2(g) → no2(g)
if these reactions come to equilibrium after combustion in an internal-combustion engine at 2000 k and 200 bar, estimate the mole fractions of no and no2 present for mole fractions of nitrogen and oxygen in the combustion products of 0.70 and 0.05.
Answers: 3
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You know the right answer?
Consider the reactions:
1/ 2 n2(g) + 1/2 o2(g) → no(g)
1/ 2 n2(g) + o2(g) → no2(g)
1/ 2 n2(g) + 1/2 o2(g) → no(g)
1/ 2 n2(g) + o2(g) → no2(g)
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